java - longest substring with exclude list of strings -

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i using this algorithm find common substring between 2 strings. please, me using array of common substrings of strings, should ignore in function.

my code in java:

public static string longestsubstring(string str1, string str2) {          stringbuilder sb = new stringbuilder();         if (str1 == null || str1.isempty() || str2 == null || str2.isempty()) {             return "";         }          // java initializes them 0         int[][] num = new int[str1.length()][str2.length()];         int maxlen = 0;         int lastsubsbegin = 0;          (int = 0; < str1.length(); i++) {             (int j = 0; j < str2.length(); j++) {                 if (str1.charat(i) == str2.charat(j)) {                     if ((i == 0) || (j == 0)) {                         num[i][j] = 1;                     } else {                         num[i][j] = 1 + num[i - 1][j - 1];                     }                      if (num[i][j] > maxlen) {                         maxlen = num[i][j];                         // generate substring str1 =>                         int thissubsbegin = - num[i][j] + 1;                         if (lastsubsbegin == thissubsbegin) {                             //if current lcs same last time block ran                             sb.append(str1.charat(i));                         } else {                             //this block resets string builder if different lcs found                             lastsubsbegin = thissubsbegin;                             sb = new stringbuilder();                             sb.append(str1.substring(lastsubsbegin, + 1));                         }                     }                 }             }         }          return sb.tostring();     }

so, function should looks like:

public static string longestsubstring(string str1, string str2, string[] ignore)

create suffix tree of 1 of strings , run through second see substring can found in suffix tree.

info on suffixtrees: http://en.wikipedia.org/wiki/suffixtree


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